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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(0,0\left(8\right)=\frac{1}{10}\cdot0,\left(8\right)=\frac{1}{10}\cdot0,\left(1\right)\cdot8=\frac{4}{5}\cdot\frac{1}{9}=\frac{4}{45}\)
\(0,1\left(2\right)=0,1+0,0\left(2\right)=\frac{1}{10}+\frac{1}{10}\cdot0,\left(2\right)=\frac{1}{10}+\frac{1}{10}\cdot0,\left(1\right)\cdot2=\frac{1}{10}+\frac{1}{5}\cdot\frac{1}{9}=\frac{1}{10}+\frac{1}{45}=\frac{11}{90}\)
\(0,1\left(23\right)=0,1+0,\left(23\right)=\frac{1}{10}+0,\left(01\right)\cdot23=\frac{1}{10}+\frac{1}{99}\cdot23=\frac{1}{10}+\frac{23}{99}=\frac{329}{990}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. $\frac{3}{-7}=\frac{-27}{63}$
$\frac{-5}{9}=\frac{-35}{63}$
Do $\frac{27}{63}< \frac{35}{63}$ nên $\frac{-27}{63}> \frac{-35}{63}$
$\Rightarrow \frac{3}{-7}> \frac{-5}{9}$
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b.
$-0,625=\frac{-625}{1000}=\frac{-5}{8}=\frac{-125}{200}$
$\frac{-19}{50}=\frac{-76}{200}> \frac{-125}{200}$
$\Rightarrow -0,625> \frac{-19}{50}$
c.
$-2\frac{5}{9}=-(2+\frac{5}{9})=\frac{-23}{9}=-(\frac{-23}{-9})$
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
C = 2-3 + (52)3.5-3 + 4-3.16 - 2.32 - 105.(\(\dfrac{24}{51}\))0
C = \(\dfrac{1}{8}\) + 56.5-3 + 4-3.42 - 2.9 - 105.1
C = \(\dfrac{1}{8}\) + 53 + \(\dfrac{1}{4}\) - 18 - 105
C = (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\)) - (105 - 125 + 18)
C = \(\dfrac{3}{8}\) - (-20 + 18)
C = \(\dfrac{3}{8}\) + 2
C = \(\dfrac{19}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ a/\(-\frac{7}{18}=\left(-\frac{7}{2}\right)\left(\frac{1}{9}\right)\)
b/\(-\frac{7}{18}=\left(-\frac{7}{9}\right):2\)
2/
a/\(\frac{7}{15}\cdot\left(-\frac{3}{8}-\frac{3}{7}\right)=\frac{7}{15}\cdot\left(-\frac{45}{56}\right)=-\frac{3}{8}\)
b/\(\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+-\frac{4}{4}\right):\frac{3}{7}\)
\(=\left(-\frac{7}{20}\right):\frac{3}{7}+\left(-\frac{2}{5}\right):\frac{3}{7}\)
\(=\left(-\frac{49}{60}\right)+\left(-\frac{14}{15}\right)=-\frac{7}{4}\)
c/\(\frac{2}{3}\cdot\left(-\frac{5}{2}\right)+\frac{10}{15}\cdot\left(-\frac{3}{7}\right)-\frac{2}{3}\cdot\left(-\frac{5}{3}\right)\)
\(=\frac{2}{3}\cdot\left(-\frac{5}{2}-\frac{3}{7}+\frac{5}{3}\right)=-\frac{53}{63}\)
3/
\(2-\left(3-x\right)=-\frac{3}{2}\)
\(2-3+x=-\frac{3}{2}\)
\(x=-\frac{3}{2}+3-2=-\frac{1}{2}\)
4/
a/ Ta có 2 trường hợp:
TH1: \(x-3,5=7,5\)
\(x=7,5+3,5=11\)
TH2: \(x-3,5=-7,5\)
\(x=-7,5+3,5=-4\)
b/ Ta có 2 trường hợp:
TH1:\(x-0,4=3,6\)
\(x=4\)
TH2: \(x-0,4=-3,6\)
\(x=-3.2\)
c/ Ta có 2 trường hợp:
TH1:\(x+\frac{4}{5}=\frac{3}{2}\)
\(x=\frac{7}{10}\)
TH2:\(x+\frac{4}{5}=-\frac{3}{2}\)
\(x=-\frac{32}{10}\)
Lời giải:
a)
\(0,(7)=\frac{7}{9}\); \(0,(23)=\frac{23}{99}\)
b) \(1,(7)=1+0,(7)=1+\frac{7}{9}=\frac{16}{9}\)
\(1,(23)=1+0,(23)=1+\frac{23}{99}=\frac{122}{99}\)
c) \(1,2(7)=\frac{12,(7)}{10}=\frac{12+0,(7)}{10}=\frac{12+\frac{7}{9}}{10}=\frac{23}{18}\)
\(1,7(23)=\frac{17,(23)}{10}=\frac{17+0,(23)}{10}=\frac{17+\frac{23}{99}}{10}=\frac{853}{495}\)